A fully loaded boeing aircraft has a mass of `5.0times10^(5)kg`.Its total wing area is `500m^(2)`.It is in level flight with a speed of `1080km/h`.If the density of air `rho` is `1.2kg m^(-3)`,the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface in percentage will be. `(g=10m/s^(2))`

Options

A. `16`
B. `8`
C. `10`
D. `6`

Answer

To solve the problem, we need to find the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface in percentage.

### Step-by-Step Solution:

1. **Convert the speed of the aircraft from km/h to m/s:**
\[ \text{Speed (v)} = 1080 \text{ km/h} = \frac{1080 \times 1000}{3600} \text{ m/s} = 300 \text{ m/s} \]

**Hint:** Remember to convert km/h to m/s by multiplying by \(\frac{1000}{3600}\).

2. **Calculate the lift force (L) using the weight of the aircraft:**
\[ L = mg = (5.0 \times 10^5 \text{ kg}) \times (10 \text{ m/s}^2) = 5.0 \times 10^6 \text{ N} \]

**Hint:** The lift force is equal to the weight of the aircraft in level flight.

3. **Use Bernoulli's equation to relate the pressure difference to the speed difference:**
The lift force can be expressed as:
\[ L = (p_1 - p_2) \cdot A \]
where \(p_1\) is the pressure on the lower surface, \(p_2\) is the pressure on the upper surface, and \(A\) is the wing area.

4. **Apply Bernoulli's principle:**
\[ p_1 + \frac{1}{2} \rho v^2 = p_2 + \frac{1}{2} \rho (v + \Delta v)^2 \]
Rearranging gives:
\[ p_1 - p_2 = \frac{1}{2} \rho \left((v + \Delta v)^2 - v^2\right) \]
Simplifying further:
\[ p_1 - p_2 = \frac{1}{2} \rho \left(2v\Delta v + \Delta v^2\right) \approx \frac{1}{2} \rho (2v\Delta v) \quad (\text{for small } \Delta v) \]

5. **Substituting into the lift equation:**
\[ L = (p_1 - p_2) \cdot A = \frac{1}{2} \rho (2v\Delta v) \cdot A \]
Setting the lift equal to the weight:
\[ 5.0 \times 10^6 = \frac{1}{2} \cdot 1.2 \cdot (2 \cdot 300 \cdot \Delta v) \cdot 500 \]

6. **Solving for \(\Delta v\):**
\[ 5.0 \times 10^6 = 1.2 \cdot 300 \cdot \Delta v \cdot 250 \]
\[ 5.0 \times 10^6 = 90000 \Delta v \]
\[ \Delta v = \frac{5.0 \times 10^6}{90000} = \frac{50000}{9} \approx 5555.56 \text{ m/s} \]

7. **Calculate the fractional increase in speed:**
\[ \text{Fractional increase} = \frac{\Delta v}{v} = \frac{5555.56}{300} \approx 18.52 \]
To express this as a percentage:
\[ \text{Percentage increase} = 18.52 \times 100 \approx 1852\% \]

### Final Answer:
The fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface is approximately **1852%**.